1
$\begingroup$

What are $y_1[n]=x_1[n]*h[n]$ and $y_2[n]=x_2[n]*h[n]$? $$x_1[n]=(0.1)^nu[n],\quad x_2[n]=(0.2)^n,\quad h[n]=(0.3)^nu[n]$$

I am confused about how to calculate $y_{1}[n]$. The formula for discrete convolution is: $$\sum_{m=-\infty}^{\infty}x[m]h[n-m] = \sum_{m=-\infty}^{\infty}h[m]x[n-m]$$.

My attempt: I chose to shift $x_{1}[n]$ along m-axis: $$y_{1}[n] = \sum_{m=0}^{n} (0.3)^{m}(0.1)^{n-m}$$ $$y_{1}[n] =(0.1)^{n}\sum_{m=0}^{n}(\frac{0.3}{0.1})^{m}$$ $$y_{1}[n] = (0.1)^{n}\sum_{m=0}^{n}(3)^{m}$$

This gives me $y_{1}[n] = \infty$. However, when I shift $h[n]$ along the m-axis I get $ y_{1}[n]$ as a finite value. My understanding of discrete time convolution was that shifting either $x[n]$ or $h[n]$ would result in the same answer.

$\endgroup$
0
$\begingroup$

You're right that both results should be identical. The mistake lies in your conclusion that $y_1[n]=\infty$ in the first case, that's not true. You get

$$y_1[n]=(0.1)^nu[n]\sum_{m=0}^n3^m=(0.1)^nu[n]\frac{3^{n+1}-1}{3-1}=\ldots\tag{1}$$

Note that you must also include the unit step $u[n]$ in $(1)$ because the response is zero for $n<0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.